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Chapter 7: Problem 894

A wire is stretched by \(0.01 \mathrm{~m}\) by a certain force \(\mathrm{F}\). Another wire of same material whose diameter and length are double to the original wire is stretched by the same force? Then what will be its elongation? (A) \(0.005 \mathrm{~m}\) (B) \(0.01 \mathrm{~m}\) (C) \(0.02 \mathrm{~m}\) (D) \(0.002 \mathrm{~m}\)

Short Answer

Expert verified
The elongation of the second wire is \(0.005 \mathrm{~m}\), which corresponds to option (A).

Step by step solution

01

Identifying relevant formulas

It's important to note that the elongation of a wire when subjected to a force depends on its dimensions, material, and the applied force. We can use the formula featuring Young's modulus (Y) for this problem: \[ Y = \frac{F\ell_1}{A_1\Delta x_1} = \frac{F\ell_2}{A_2\Delta x_2} \] where: - \(Y\) is the Young's modulus for the material - \(F\) is the force applied - \(\ell_1\) and \(\ell_2\) represent the initial lengths of the two wires - \(A_1\) and \(A_2\) represent the cross-sectional areas of the two wires - \(\Delta x_1\) and \(\Delta x_2\) are the corresponding elongations of the two wires when the force is applied.
02

Simplify equation using given conditions

According to the problem, the length and the diameter of the second wire are double that of the first wire. We can denote these relationships as follows: \(\ell_2 = 2\ell_1\) and \(d_2 = 2d_1\). Considering the area of a circle, the cross-sectional area for the two wires is \(A_1 = \pi (d_1/2)^2\) and \(A_2 = \pi (d_2/2)^2\). Now replace \(d_2\) with \(2d_1\) as given in the problem and re-write the equation for \(A_2\): \[ A_2 = \pi (2d_1/2)^2 = 4\pi (d_1/2)^2 \] Therefore, \(A_2 = 4A_1\). Now, let's substitute these ratios in a simplified form of our main equation: \[ \frac{F\ell_1}{A_1\Delta x_1} = \frac{F(2\ell_1)}{4A_1\Delta x_2} \]
03

Solve for elongation of the second wire

We have to solve the equation for \(\Delta x_2\): \[ \frac{F\ell_1}{A_1\Delta x_1} = \frac{2F\ell_1}{4A_1\Delta x_2} \] Now, notice that \(F\), \(\ell_1\), and \(A_1\) are the same in both terms, so we can remove them from the equation. \[ \frac{1}{\Delta x_1} = \frac{2}{4\Delta x_2} \] Thus, \[ \Delta x_2 = \frac{1}{2} \Delta x_1 = 0.5 \times 0.01 ~\text{m} = 0.005 ~\text{m} \] The elongation of the second wire is 0.005 m, which corresponds to option (A).

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